How to apply function by groups in array in R? -

i have 3d array latitude, longitude , datetime (year_month_day_hour). best way in r apply function on array groups (in case years or month or days)? result should array mean values. 3 dimension year, month or day.

str(data)  num [1:7, 1:7, 1:5] 977 994 1010 1020 1026 ...  - attr(*, "dimnames")=list of 3   ..$ : chr [1:7] "60" "57.5" "55" "52.5" ...   ..$ : chr [1:7] "-30" "-27.5" "-25" "-22.5" ...   ..$ : chr [1:5] "2014_10_01_00" "2014_10_01_06" "2014_10_01_12" "2014_10_01_18" ... 

example (truncated):

dput(data) structure(c(977.2, 994.4, 1009.8, 1020.1, 1026.4, 1029.4, 1029.2,  978.7, 995.7, 1010.2, 1020.5, 1026.5, 1028.8, 1028.3, 982, 997.5,  1011.3, 1021.2, 1026.1, 1027.4, 1027.1, 986.2, 999.9, 1013, 1021.7,  1025.1, 1025.7, 1026, 990.6, 1002.7, 1014.5, 1021.3, 1023.9,  1024.7, 1025.6, 995.1, 1005.7, 1015.2, 1019.9, 1022.6, 1024.5,  1025.9, 999.1, 1008, 1015.1, 1018.6, 1021.8, 1024.5, 1026.6,  982.1, 998.9, 1011.8, 1020.1, 1025.5, 1028.4, 1028.8, 981.9,  999.3, 1012.7, 1021.2, 1026.4, 1028.8, 1029, 983.9, 1000.2, 1013.5,  1022.1, 1027, 1028.9, 1028.9, 987.1, 1001.8, 1014.6, 1022.7,  1027.3, 1028.6, 1028.2, 990.9, 1004.1, 1016.1, 1023.3, 1027.2,  1027.9, 1027.4, 995.1, 1006.9, 1017.8, 1023.8, 1026.8, 1027,  1026.9, 999.5, 1010.1, 1019.1, 1023.8, 1025.9, 1026.1, 1026.9,  990.3, 1002.3, 1010.9, 1018.3, 1024, 1027.6, 1028.6, 990.6, 1004.1,  1013.2, 1020.8, 1026.2, 1029.3, 1029.8, 992.1, 1005.5, 1015.2,  1023, 1028, 1030.4, 1030.5, 994.5, 1007, 1017.2, 1024.7, 1029.4,  1031, 1030.3, 997.4, 1008.8, 1019, 1025.7, 1030, 1031, 1029.8,  1000.1, 1010.9, 1020.9, 1026.5, 1030, 1030.6, 1029.5, 1002.9,  1013.3, 1022.6, 1027.2, 1029.7, 1029.7, 1029.2, 993.6, 997.5,  1001.3, 1007.4, 1015.5, 1022.7, 1026.4, 996.1, 1001.1, 1005.8,  1012.7, 1020.1, 1025.6, 1027.9, 998.4, 1004.5, 1010.4, 1017.6,  1023.8, 1027.6, 1029.1, 1000.2, 1007.3, 1014.4, 1021.5, 1026.4,  1029, 1029.7, 1002, 1010, 1017.8, 1024.3, 1028.4, 1029.9, 1029.6,  1004.3, 1012.9, 1020.7, 1026.3, 1029.7, 1030.2, 1029.3, 1006.9,  1016, 1023.2, 1027.7, 1030.3, 1029.7, 1028.6, 987.9, 989.6, 995.1,  1002.9, 1010.8, 1018.9, 1025.1, 989.8, 990, 995.1, 1004.7, 1013.9,  1021.8, 1026.8, 993.1, 992.6, 998.1, 1008.8, 1018, 1024.6, 1028.3,  996.9, 997.3, 1003.9, 1014, 1021.9, 1026.8, 1029.1, 1000.3, 1003.1,  1010.5, 1019, 1025.2, 1028.5, 1029.6, 1003.6, 1008.7, 1016.4,  1023.1, 1027.8, 1029.8, 1029.9, 1007.3, 1013.7, 1020.8, 1026.3,  1029.8, 1030.2, 1029.6), .dim = c(7l, 7l, 5l), .dimnames = list( c("60", "57.5", "55", "52.5", "50", "47.5", "45"), c("-30",  "-27.5", "-25", "-22.5", "-20", "-17.5", "-15"), c("2014_10_01_00",  "2014_10_01_06", "2014_10_01_12", "2014_10_01_18", "2014_10_02_00" ))) 

solution:

group <- as.factor(as.date(dimnames(data)[[3]],format="%y_%m_%d"))  aperm(apply(data,c(1,2), by, group, mean),c(2,3,1)) 

just specify dimension second argument in apply function. example summing "date" margin:

> apply(array, 3, sum) # 2014_10_01_00 2014_10_01_06 2014_10_01_12 2014_10_01_18 2014_10_02_00  #       49691.3       49782.3       49919.6       49851.4       49639.0  

if dimensions have names can use name character string second argument.

---

edit

op wants results grouped date. function can maybe give guidance desired result:

myapply <- function(array, d, fun){   # function apply "fun" "array"   # of class array dimension   # 3. array grouped d number   # between 1 , 4   # 1: year   # 2: month   # 3: day   # 4: hour   d.name <- strsplit(dimnames(array)[[3]], "_")    # make groups   names  <- lapply(d.name, function(x, d)                       paste(x[1:d], collapse= "_"), d = d)   groups <- unique(names)    # indices groups   indices <- lapply(groups, function(x, names)                        which(unlist(names) %in% x), names = names)    # compute function on groups   results <- lapply(indices, function(ind, arr, fun)                         fun(as.vector(arr[,,ind])), arr = array, fun = fun)    names(results) <- unlist(groups)   return(results) } 

results:

# mean grouping day myapply(array, 3, mean) # $`2014_10_01` # [1] 1016.554 #  # $`2014_10_02` # [1] 1013.041   # mean, grouping hour myapply(array, 4, mean) # $`2014_10_01_00` # [1] 1014.108 #  # $`2014_10_01_06` # [1] 1015.965 #  # $`2014_10_01_12` # [1] 1018.767 #  # $`2014_10_01_18` # [1] 1017.376 #  # $`2014_10_02_00` # [1] 1013.041